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        <section id="main"><article id="post-sort/快速排序(Quicksort)" class="article article-type-post" itemscope itemprop="blogPost">
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      快速排序(Quicksort)
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        <p><a href="http://www.runoob.com/w3cnote/sort-algorithm-summary.html" target="_blank" rel="noopener">http://www.runoob.com/w3cnote/sort-algorithm-summary.html</a></p>
<table>
<thead>
<tr>
<th>排序算法</th>
<th>时间复杂度</th>
</tr>
</thead>
<tbody>
<tr>
<td>冒泡排序</td>
<td>O(n2)</td>
</tr>
<tr>
<td>选择排序</td>
<td>O(n2)</td>
</tr>
<tr>
<td>插入排序</td>
<td>O(n2)</td>
</tr>
<tr>
<td>希尔排序</td>
<td>O(n1.5)</td>
</tr>
<tr>
<td>快速排序</td>
<td>O(N*logN)</td>
</tr>
<tr>
<td>归并排序</td>
<td>O(N*logN)</td>
</tr>
<tr>
<td>堆排序</td>
<td>O(N*logN)</td>
</tr>
<tr>
<td>基数排序</td>
<td>O(d(n+r))</td>
</tr>
</tbody>
</table>
<h5 id="快速排序-Quicksort"><a href="#快速排序-Quicksort" class="headerlink" title="快速排序(Quicksort)"></a>快速排序(Quicksort)</h5><ul>
<li>基本思想：（分治）</li>
</ul>
<p>先从数列中取出一个数作为key值；<br>将比这个数小的数全部放在它的左边，大于或等于它的数全部放在它的右边；<br>对左右两个小数列重复第二步，直至各区间只有1个数。</p>
<ul>
<li>辅助理解：<a href="https://blog.csdn.net/morewindows/article/details/6684558/" target="_blank" rel="noopener">挖坑填数</a></li>
</ul>
<ol>
<li>初始时 i = 0; j = 9; key=72<br>由于已经将a[0]中的数保存到key中，可以理解成在数组a[0]上挖了个坑，可以将其它数据填充到这来。<br>从j开始向前找一个比key小的数。当j=8，符合条件，a[0] = a[8] ; i++ ; 将a[8]挖出再填到上一个坑a[0]中。<br>这样一个坑a[0]就被搞定了，但又形成了一个新坑a[8]，这怎么办了？简单，再找数字来填a[8]这个坑。<br>这次从i开始向后找一个大于key的数，当i=3，符合条件，a[8] = a[3] ; j– ; 将a[3]挖出再填到上一个坑中。</li>
</ol>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">数组：72 - 6 - 57 - 88 - 60 - 42 - 83 - 73 - 48 - 85</span><br><span class="line"> 0   1   2    3    4    5    6    7    8    9</span><br></pre></td></tr></table></figure>
<ol>
<li>此时 i = 3; j = 7; key=72<br>再重复上面的步骤，先从后向前找，再从前向后找。<br>从j开始向前找，当j=5，符合条件，将a[5]挖出填到上一个坑中，a[3] = a[5]; i++;<br>从i开始向后找，当i=5时，由于i==j退出。<br>此时，i = j = 5，而a[5]刚好又是上次挖的坑，因此将key填入a[5]。</li>
</ol>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">数组：48 - 6 - 57 - 88 - 60 - 42 - 83 - 73 - 88 - 85</span><br><span class="line"> 0   1   2    3    4    5    6    7    8    9</span><br></pre></td></tr></table></figure>
<ol>
<li>可以看出a[5]前面的数字都小于它，a[5]后面的数字都大于它。因此再对a[0…4]和a[6…9]这二个子区间重复上述步骤就可以了。</li>
</ol>
<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br></pre></td><td class="code"><pre><span class="line">&lt;数组：48 - 6 - 57 - 42 - 60 - 72 - 83 - 73 - 88 - 85</span><br><span class="line"> 0   1   2    3    4    5    6    7    8    9</span><br></pre></td></tr></table></figure>
<ul>
<li><p>平均时间复杂度：O(N*logN)</p>
</li>
<li><p>方式一</p>
</li>
</ul>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span> <span class="class"><span class="keyword">class</span> <span class="title">Quick</span> </span>&#123;</span><br><span class="line"></span><br><span class="line">   <span class="function"><span class="keyword">public</span> <span class="keyword">static</span> <span class="keyword">void</span> <span class="title">sort</span><span class="params">(<span class="keyword">int</span> arr[], <span class="keyword">int</span> low, <span class="keyword">int</span> high)</span> </span>&#123;</span><br><span class="line">       <span class="keyword">int</span> l = low;</span><br><span class="line">       <span class="keyword">int</span> h = high;</span><br><span class="line">       <span class="keyword">int</span> povit = arr[low];</span><br><span class="line"></span><br><span class="line">       <span class="keyword">while</span> (l &lt; h) &#123;</span><br><span class="line">           <span class="keyword">while</span> (l &lt; h &amp;&amp; arr[h] &gt;= povit)</span><br><span class="line">               h--;</span><br><span class="line">           <span class="keyword">if</span> (l &lt; h) &#123;</span><br><span class="line">               <span class="keyword">int</span> temp = arr[h];</span><br><span class="line">               arr[h] = arr[l];</span><br><span class="line">               arr[l] = temp;</span><br><span class="line">               l++;</span><br><span class="line">           &#125;</span><br><span class="line">           <span class="keyword">while</span> (l &lt; h &amp;&amp; arr[l] &lt;= povit)</span><br><span class="line">               l++;</span><br><span class="line">           <span class="keyword">if</span> (l &lt; h) &#123;</span><br><span class="line">               <span class="keyword">int</span> temp = arr[h];</span><br><span class="line">               arr[h] = arr[l];</span><br><span class="line">               arr[l] = temp;</span><br><span class="line">               h--;</span><br><span class="line">           &#125;</span><br><span class="line">       &#125;</span><br><span class="line">       <span class="comment">//打印</span></span><br><span class="line">       <span class="keyword">for</span> (<span class="keyword">int</span> i = <span class="number">0</span>; i &lt; arr.length; i++) &#123;</span><br><span class="line">           System.out.print(arr[i]);</span><br><span class="line">       &#125;</span><br><span class="line">       System.out.println();</span><br><span class="line"></span><br><span class="line">       <span class="comment">//       print(arr);</span></span><br><span class="line">       System.out.println(<span class="string">"l="</span> + (l + <span class="number">1</span>) + <span class="string">"h="</span> + (h + <span class="number">1</span>) + <span class="string">"povit="</span> + povit + <span class="string">"\n"</span>);</span><br><span class="line">       <span class="keyword">if</span> (l &gt; low) sort(arr, low, l - <span class="number">1</span>);</span><br><span class="line">       <span class="keyword">if</span> (h &lt; high) sort(arr, l + <span class="number">1</span>, high);</span><br><span class="line">   &#125;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>
<ul>
<li>方式二 runoob</li>
</ul>
<figure class="highlight java"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br><span class="line">28</span><br><span class="line">29</span><br><span class="line">30</span><br><span class="line">31</span><br><span class="line">32</span><br><span class="line">33</span><br><span class="line">34</span><br><span class="line">35</span><br><span class="line">36</span><br><span class="line">37</span><br><span class="line">38</span><br><span class="line">39</span><br><span class="line">40</span><br><span class="line">41</span><br><span class="line">42</span><br><span class="line">43</span><br><span class="line">44</span><br><span class="line">45</span><br><span class="line">46</span><br><span class="line">47</span><br></pre></td><td class="code"><pre><span class="line"><span class="keyword">public</span>  &lt;T extends Comparable&lt;? <span class="keyword">super</span> T&gt;&gt; T[] quickSort(T[] targetArr, <span class="keyword">int</span> start, <span class="keyword">int</span> end) &#123;</span><br><span class="line"></span><br><span class="line">      <span class="keyword">int</span> i = start + <span class="number">1</span>;</span><br><span class="line">      <span class="keyword">int</span> j = end;</span><br><span class="line">      T key = targetArr[start];</span><br><span class="line">      <span class="keyword">while</span> (i &lt; j) &#123;</span><br><span class="line">          <span class="comment">/**</span></span><br><span class="line"><span class="comment">           * 按照j--的方向遍历目标数组，直到比key小的值为止</span></span><br><span class="line"><span class="comment">           */</span></span><br><span class="line">          <span class="keyword">while</span> (j &gt; i &amp;&amp; targetArr[j].compareTo(key) &gt;= <span class="number">0</span>) &#123;</span><br><span class="line">              j--;</span><br><span class="line">          &#125;</span><br><span class="line">          <span class="keyword">if</span> (i &lt; j) &#123;</span><br><span class="line">              <span class="comment">/**</span></span><br><span class="line"><span class="comment">               * targetArr[i]已经保存在key中，可将后面的数填入</span></span><br><span class="line"><span class="comment">               */</span></span><br><span class="line">              targetArr[i] = targetArr[j];</span><br><span class="line">              i++;</span><br><span class="line">          &#125;</span><br><span class="line">          <span class="comment">/**</span></span><br><span class="line"><span class="comment">           * 按照i++的方向遍历目标数组，直到比key大的值为止</span></span><br><span class="line"><span class="comment">           */</span></span><br><span class="line">          <span class="keyword">while</span> (i &lt; j &amp;&amp; targetArr[i].compareTo(key) &lt;= <span class="number">0</span>) &#123;</span><br><span class="line">              <span class="comment">/**</span></span><br><span class="line"><span class="comment">               * 此处一定要小于等于零，假设数组之内有一亿个1，0交替出现的话，而key的值又恰巧是1的话，那么这个小于等于的作用就会使下面的if语句少执行一亿次。</span></span><br><span class="line"><span class="comment">               */</span></span><br><span class="line">              i++;</span><br><span class="line">          &#125;</span><br><span class="line">          <span class="keyword">if</span> (i &lt; j) &#123;</span><br><span class="line">              <span class="comment">/**</span></span><br><span class="line"><span class="comment">               * targetArr[j]已保存在targetArr[i]中，可将前面的值填入</span></span><br><span class="line"><span class="comment">               */</span></span><br><span class="line">              targetArr[j] = targetArr[i];</span><br><span class="line">              j--;</span><br><span class="line">          &#125;</span><br><span class="line">      &#125;</span><br><span class="line">      <span class="comment">/**</span></span><br><span class="line"><span class="comment">       * 此时i==j</span></span><br><span class="line"><span class="comment">       */</span></span><br><span class="line">      targetArr[i] = key;</span><br><span class="line">      <span class="comment">//递归调用，把key前面的完成排序</span></span><br><span class="line">      <span class="keyword">this</span>.quickSort(targetArr, start, i - <span class="number">1</span>);</span><br><span class="line"></span><br><span class="line">      <span class="comment">//递归调用，把key后面的完成排序</span></span><br><span class="line">      <span class="keyword">this</span>.quickSort(targetArr, j + <span class="number">1</span>, end);</span><br><span class="line">      <span class="keyword">return</span> <span class="keyword">null</span>;</span><br><span class="line">  &#125;</span><br></pre></td></tr></table></figure>
      
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